# Why isn't there a contradiction between the existence of CNOT gate/entanglement and the no-cloning theorem?

Of course I am not implying that I am right and the no cloning theorem is wrong, but I am trying to figure out what is wrong with my reasoning and yet I couldn't find the mistake.

Based on Wikipedia

In physics, the no-cloning theorem states that it is impossible to create an identical copy of an arbitrary unknown quantum state.

So we start with a standard qubit $$|\psi\rangle$$ with completely unknown state where:

$$|\psi \rangle = \alpha |0\rangle + \beta|1\rangle$$

That qubit has probability $$\alpha^2$$ of being $$0$$ and probability $$\beta^2$$ of being 1 and if I understand the theory correctly it will not be possible to duplicate this qubit without knowing both $$\alpha$$ and $$\beta$$.

Now, if we plug this qubit along with a $$|0\rangle$$ into a $$CNOT$$ it seems to me that we end up with 2 identical qubits, each of them has probability $$\alpha^2$$ of being $$0$$ and probability $$\beta^2$$ of being 1.

Here is the math:

$$CNOT |\psi, 0\rangle = \\ CNOT [(\alpha |0\rangle + \beta|1\rangle)\otimes |0\rangle] = \\ \begin{bmatrix}1&&0&&0&&0\\0&&1&&0&&0\\0&&0&&0&&1\\0&&0&&1&&0\end{bmatrix} \begin{pmatrix}\alpha\\\beta\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} =\\ \begin{bmatrix}1&&0&&0&&0\\0&&1&&0&&0\\0&&0&&0&&1\\0&&0&&1&&0\end{bmatrix} \begin{pmatrix}\alpha\\0\\\beta\\0\end{pmatrix} =\\ \begin{pmatrix}\alpha\\0\\0\\\beta\end{pmatrix} = \alpha \begin{pmatrix}1\\0\\0\\0\end{pmatrix}+\beta \begin{pmatrix}0\\0\\0\\1\end{pmatrix}=\\ \alpha |00\rangle + \beta |11\rangle$$ So the result becomes 2 exactly identical qubits, both have identical probabilities of being zero and identical probabilities of being one.

Since I am sure that the no-cloning theorem can't be wrong, I am asking what is wrong with my reasoning.

• Unfortunately the no cloning theorem is often stated informally and "in words", the precise statement in my opinion is much easier to understand: given a state $|\psi\rangle$ and a blank state $|0\rangle$ independent of $|\psi\rangle$, there exists no unitary $U$ such that $U|\psi\rangle|0\rangle=e^{i\alpha}|\psi\rangle|\psi\rangle$ for some phase $\alpha$. This is clearly not what the CNOT does, as Mariia Mykhailova explains in her answer. – user2723984 Jul 31 '19 at 9:07

The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $$\big(\alpha |0\rangle + \beta |1\rangle \big) \otimes \big(\alpha |0\rangle + \beta |1\rangle \big)$$. This is not the state CNOT will give you.